Answer:
a) 1.9%
b) T2s = 505.5 k = 232.5°C
Explanation:
P1 = 95 kPa
T1 = 27°C = 300 k
P2 = 600 kPa
T1 = 277°c = 550 k
Table used : Table ( A - 17 ) Ideal gas properties of air
<u>a) determining the isentropic efficiency of the compressor </u>
Л = ( h2s - h1 ) / ( h2a - h1 ) ---- ( 1 )
where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k
To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)
Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306 hence; h2s = 500.72 kJ/kg
back to equation1
Л = 0.019 = 1.9%
<u>b) Calculate the exit temperature of the air if compressor is reversible </u>
if compressor is reversible the corresponding exit temperature
T2s = 505.5 k = 232.5°C
given that h2s = 500.72 kJ/kg
Answer:
total_cost = cost + tax
Explanation:
Step1) Let the 2 variable for take input from user e.g price and quantity
var price ;
var quantity ;
var cost ;
var tax ;
var total_cost ;
Step2) take input from user quantity of item 'A'
step3) cost = price * quantity
Step4) tax = 0.08 * cost
Step5) total_cost = cost + tax
Step6) print the total_cost
Answer:
The distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2
Explanation:
From the image uploaded, a Face centered cubic structure (100) plane, there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell.
In terms of the atomic radius, R, we determine the distance between the centers of adjacent atoms.
Let this distance = AC
the two adjacent sides = AB and BC
AB = a = 2R
BC = a = 2R
Using Pythagoras theorem
AC² = AB² + BC²
AC² = a² + a²
AC² = 2a²
AC = √2a²
AC = a√2
But a = 2R
AC = 2R√2
Therefore, the distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2
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